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From: Henno Brandsma
Date: Jan 1, 2005
Subject: Alexander's Subbase Lemma and applications

Recall the definition of a subbase for a space X: a collection S of open subsets
of X is called a subbase for X when the finite intersections from S form a base
for the topology of X. Many topologies, like the initial topologies (see the note on initial
topologies; product topologies are a special case) and
the order topology on a linearly ordered set are most naturally defined by subbases. Their main
use as a topological tool is the following classical theorem, modestly called a "lemma":

Alexander Subbase Lemma.
Let S be a subbase for a topological space X.
Then X is compact iff every cover of X by elements from S has a finite subcover.

Proof.
The direction from left to right is trivial: a cover by elements of S is just a
special type of open cover, and so by compactness \emph{every} open cover has a finite
subcover.
For the other direction we need a version of the Axiom of Choice. We will use
Zorn's lemma: a partially ordered set P in which every chain has an upperbound,
has a maximal element.
Assume every cover from S has a finite subcover, and X is not compact.
Then X has an open cover without a finite subcover, and by Zorn's lemma there is a maximal
(with respect to inclusion) open cover without a finite subcover.

(If you see this right away, skip forward a bit, this is quite standard)
Check for aplication of Zorn's lemma: Let the partially ordered set P be the set of all
covers of X without a finite subcover, ordered by inclusion.
This is of course a partial order, and non-empty by assumption.
Let C be a non-empty chain in P. So C is a set of open covers in P, such that for any two
U_1 and U_2 in C, we have U_1 \subset U_2 or U_2 \subset U_1.
Let U be the union of all these covers. This is surely an upperbound wrt inclusion
of all U_i in C, if only U were in P, that is, U is a cover without a finite subcover.

So suppose that U has a finite subcover O_1,...,O_n, all in U = \/{V: V in C}.
So for each i from {1,...,n} we have V_i from C, such that O_i is in V_i.
The V_i (i=1,..,n) have a maximum wrt inclusion as the V_i are in C, which is by
assumption linearly ordered by inclusion; call this maximum V_j.
Then all O_i \in V_i \subset V_j (i=1..n) and so all O_i are in V_j and the O_i
cover X and so V_j would have a finite subcover, which it cannot have.
So U has no finite subcover, and thus is an upperbound for the chain C and
Zorn's lemma applies. The U from our main proof is ust the promised maximal element.
--

Properties of U:
1) We have the following: if O is not in U then U \/ {O} has a finite
subcover (or this larger cover would contradict maximality of U) and this cover must
include O (or U would already have had a finite subcover), so there are U_1,...,U_n in U
such that X = \/_{i=1..n} U_i \/ O.

2) The converse also holds: if there are such finitely many U_i from U that together
with O cover X, then O cannot be in U (or we would again have a finite subcover of U).

3) If V_1, V_2,...,V_n are all not in U, then so is V_1 /\ .... /\ V_n.
For we have for each i=1,..,n subsets V_{i,j} (j=1,..,n_i) such that
V_i \/ \/{j=1..n_i} V_{i,j} = X.
But then x in X is either in some V_{i.j}, or (if not!) it is in all V_i (i=1,..,n).
So (V_1 /\ V_2 /\ .... /\ V_n) \/ \/_{i=1..n, j=1..n_i} V_{i,j} = X.
As all V_{i,j} are from U, we see from 2) that V_1 /\ ... /\ V_n is not in U.

4) If V is not in U, and V \subset W, then W is also not in V.
Use 1): there are V_1,..,V_n that together with V, cover X.
But these V_i then also cover X, if we use the larger set W instead of V.
So by 2), W is also not in U.

Aside:
Note that 3 and 4 say that the sets that are not in U form a filter of open sets.
It is also easy to show that U must be closed under finite unions, and under subsets
(so O open and O \subset O' , O' in U then O is in U too). This is because adding open subsets
and finite unions doesn't add a finite subcover if we didn't have one in the first place.
(this says that U is a so-called ideal of open subsets).

Now, we didn't use the subbase at all. In fact all of the facts about U hold in any
non-compact space (if we assume the Axiom of Choice to get U in the first place).
Claim: S /\ U covers X as well.
Proof of claim: Let x be in X. Then x is in some O from U (U is a cover), and as S is a subbase
we see that there are finitely many S_1,..,S_n from S such that
x \in /\S_i \subset O.
Now, some S_i is in U, for otherwise by 3) we have /\S_i not in U and by 4) we'd get
O not in U, which is false. So some S_i from S is in U, and x is then covered by this S_i.

Now we are done: the elements from S /\ U are now a cover by elements from S, and have a finite
subcover, which is then also a finite subcover of U, which contradicts the assumption
that U didn't have one.
So X is in fact compact, and we are done.
----
----
This allows for some nice applications, the main one of which is:

Tychonov's theorem.
Let X_i be a family of compact spaces and let X be their Cartesian product, in the
product topology. Then X is compact.

Proof:
The product topology is the so-called initial topology wrt the projections
p_i : X ---> X_i, which just means that the collection
S = {(p_i)^{-1}[O]: i in I, O open in X_i } is a subbase for the product topology.
By Alexander's Subbase Lemma, we only need to check that every cover by elements from S
has a finite subcover. So let U be a cover of X by elements from S.

Every O in U is then of the form O = (p_i)^{-1}[V]; we denote V by V(O), i by i(O).

Claim: there is some i in I such that {V(O): O in U, i(O) = i} covers X_i.
Proof: suppose not. Then for every i in I, we can pick a point x_i such that
x_i is not in any V(O) (O in U) with i(O) = i\footnote{this uses the Axiom of Choice}.
Let x be the point of X defined by these x_i (i in I).
Then for all O in U, x is not in O: if x is in O, let i = i(O) and V = V(O), and
then p_i(x) = x_i is in V(O), which is was chosen not to be in.
So U is then not a cover of X, contradiction.
This proves the claim.

Fix this i from the claim from now on.
The cover {V(O): O in U, i(O) = i} covers the compact space X_i, so has a finite subcover,
determined by O_1,...,O_n. Then these O_i cover X: let x be in X, p_i(x) is in some V(O_i)
and thus x is in (p_i)^{-1}[V(O_i)] = O_i.

So U has a finite subcover and X is compact.
---

Another application:

Compactness in ordered spaces.
Let X be an ordered space. Then X is compact iff
every subset of X has a supremum (least upper bound).

Proof. Let X satisfy the lub property.
An ordered space has a standard subbase S = {L_x: x in X} \/ {U_x: x in X}
with L_x = {y in X: y < x} and U_x = {y in X: y > x}.
To prove compactness we only need to consider covers by sets of the type L_x
and U_x, by the Alexander subbase lemma.
So let U be such a cover.

First note that in particular X has a supremum, which must be the maximum (say M) of X.
This M cannot be covered by a set of the form L_x, so U contains at least one set
of the form U_x. Also, the empty set has a supremum, and as all x in X are upperbounds
for the empty set we see that this supremum is in fact the minimum (say m) of X, which we now
see exists. This m is not covered by any U_x so U also contains at least one set of the form
L_x.

So define A = {x: L_x is in U}, and let a = sup(A), which exists by assumption.
This a is not covered by any L_x from U (or this x would be in A and larger than a)
so this is covered by some U_x (in U).
So a \in U_x, which means that x < a. As a is the \emph{least} upperbound of A, we see
that x cannot be an upperbound for A, and so for some y in A we have x < y (<= a).
But L_y is in U (as y is in A) and then U_x and L_y together cover X, as x < y.
So we have the required finite subcover of U, and by the subbase lemma, X is compact.
-

Suppose now that X is compact. Suppose that there is a subset A of X without a supremum.
If A is empty, this means that X has no minimum (see the first part of the proof) and
then the cover U_x (x in X) of X has no finite subcover: suppose U_{x_1},...,U_{x_n}
is a finite subcover, and let x_j = min(x_i : i =1,..,n), then X = \/_i U_{x_i} = U_{x_j}
and so for all x in X we would have x > x_j, which cannot be as even x = x_j contradicts this.
So suppose that A is non-empty.
If A has no upperbound, then {L_x: x in A} covers X (x in X , then x is no upperbound of A
by assumption, so some a in A is larger than x and then x is in L_a, for this a)
and thus has a finite subcover L_{x_1},...,L_{x_n}. But then setting x_j = max(x_i: i =1,..,n)
we see that X = L_{x_j}, which is contradicted by x_j.
So A does have upperbounds, let the set of upperbounds of A be U.
Now U := {L_a: a in A } \/ {U_b: b in U} covers X
(if x is in X, and x is not in any L_a (a in A), then x is not below any element of A, so x
is an upperbound, and as there is no least upperbound, there is some b in U
such that b < x, and then x is in U_b).

But if L_{a_1},...,L_{a_n} together with U_{b_1},...,U_{b_m} (a_i in A, b_i in U)
cover X then as before L_a and U_b (where a = max(a_i) and b = min(b_i)) cover X as well,
which means that a is in U_b (as it is not in L_a) and so b < a, contradicting that b
is in U = the set of upperbounds of A.
So assuming there is no sup for A, we have found a cover U without a finite subcover
and we have a contradiction with the compactness of X, and so all subsets A of X have a
supremum and this concludes the proof of the other direction.
---
---

Note: a space X with a subbase such that every cover by elements of S has a subcover
of 2 elements, is called supercompact. The Alexander Subbase Lemma implies that a
supercompact space is supercompact.
Also, one easily adapts our proof of the Tychonov theorem to see that products of
supercompact spaces are supercompact.
(as we can consider the (p_i)^{-1}[S] (S in a subbase of X_i) and use that as a subbase
for the product topology as well!).
But this property is not preserved by continuous images. Also, (harder): every compact metric
space is supercompact.
Implicit in the previous proof (hence these remarks): all ordered compact spaces are supercompact.

Henno Brandsma